Instiki
Sandbox

Markdown+itex2MML Sandbox

Play around below. Your changes will, periodically, be rolled back.

Some examples

(1)minw hp h+w rp r+w lp l\mathop{min} w_h p_h + w_r p_r + w_l p_l

Here’s an equation

(2) e x 2/2dx=2π\int_{-\infty}^\infty e^{-x^2/2} \mathrm{d}x = \sqrt{2\pi}

which we can later refer1 back to as (2).

The Dirac equation:

(iD+m)ψ=0(i\slash{D}+m)\psi =0

Here’s the table of Clifford2 algebras over :

j012345678
𝒞 j (2)(4)(8)(8)(8)(16)
𝒞 j +(2)(2)(2)(2)(2)(4)(8)(16)

where the generators of 𝒞 j ± satisfy

γ iγ j+γ jγ i=±2δ ij\gamma_i\gamma_j +\gamma_j \gamma_i =\pm 2\delta_{i j}

and 𝒞 n+8 ±=𝒞 n ±(16).

(3)lim n k=1 n1k 2=π 26\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{k^2} = \frac{\pi^2}{6}

More Examples

(4)×E=Bt\nabla \times \vec{E} = - \frac {\partial \vec{B}}{\partial t}
(5)Bdl=μ 0I enc\oint \mathbf{B}\cdot \mathrm{d}\mathbf{l} = \mu_0 I_\text{enc}

H 1(𝒵,𝒪(k)) Let G=(V,E) be a graph, with w:V[0,1] a weight function.

(6){Q i,Q j}=δ ij.\{Q_i, Q_j\} = \delta_{ij}\mathcal{H}.

Here is a groupoid 𝒢. 3(mod5).

Theorems

Definition

Let H be a subgroup of a group G. A left coset of H in G is a subset of G that is of the form xH, where xG and xH={xh:hH}.

Similarly a right coset of H in G is a subset of G that is of the form Hx, where Hx={hx:hH}.

Lemma

Let H be a subgroup of a group G, and let x and y be elements of G. Suppose that xHyH is non-empty. Then xH=yH.

Proof

Let z be some element of xHyH. Then z=xa for some aH, and z=yb for some bH. If h is any element of H then ahH and a 1hH, since H is a subgroup of G. But zh=x(ah) and xh=z(a 1h) for all hH. Therefore zHxH and xHzH, and thus xH=zH. Similarly yH=zH, and thus xH=yH, as required.

Lemma

Let H be a finite subgroup of a group G. Then each left coset of H in G has the same number of elements as H.

Proof

Let H={h 1,h 2,,h m}, where h 1,h 2,,h m are distinct, and let x be an element of G. Then the left coset xH consists of the elements xh j for j=1,2,,m. Suppose that j and k are integers between 1 and m for which xh j=xh k. Then h j=x 1(xh j)=x 1(xh k)=h k, and thus j=k, since h 1,h 2,,h m are distinct. It follows that the elements xh 1,xh 2,,xh m are distinct. We conclude that the subgroup H and the left coset xH both have m elements, as required.

Theorem

(Lagrange’s Theorem). Let G be a finite group, and let H be a subgroup of G. Then the order of H divides the order of G.

Proof

Each element x of G belongs to at least one left coset of H in G (namely the coset xH), and no element can belong to two distinct left cosets of H in G (see Lemma 1). Therefore every element of G belongs to exactly one left coset of H. Moreover each left coset of H contains H elements (Lemma 2). Therefore G=nH, where n is the number of left cosets of H in G. The result follows.

Corollary

Let x be an element of a finite group G. Then the order of x divides the order of G.

Problems?

Maybe the notation should be different than the Latex counterparts, but these do not seem to be rendering correctly (JD: They look fine to me. A font problem in your browser?) (JB: That was exactly the problem, thanks for the suggestion. The braces weren’t stretching correctly with my old fonts.):

  • SVG:
Box diagram d s¯ u, c, t s d¯ u, c, t W W+
K 0K¯ 0 Mixing
  • Complicated commutative diagrams (equations in SVG)
Complicated commutative diagram, realized in SVG 1 1 1 1 Id Id A B ρ H H K K ϕ 1 ϕ 2 N A N B N A N B
  • SVG in equations.

In SU(3), Rank-2 Symmetric Tensor Representation Fundamental Representation = Adjoint Representation Rank-3 Symmetric Tensor Representation .

  • Cases:
r a+1={0 with prob.exp(θr a) max{δr a,z} with prob.1exp(θr a) r_{a+1} = \begin{cases} 0 & \text{with prob.}\quad \exp(-\theta r_a) \\ \max \lbrace \delta r_a, z \rbrace & \text{with prob.}\quad 1 - \exp(-\theta r_a) \\ \end{cases}
  • Stretchy Brackets:
q a(z)=σ a 1exp[γ+zσ a]q_a(z) = \sigma_a^{-1} \exp{\left[ -\frac{\gamma + z}{\sigma_a} \right]}
  • Linearity of Quadrature Rules
i=1 N(αf(x i)+βg(x i))w i=α i=1 Nf(x i)w i+β i=1 Ng(x i)w i\sum_{i = 1}^N {\left( {\alpha f(x_i ) + \beta g(x_i )} \right)w_i } = \alpha \sum_{i = 1}^N {f(x_i )w_i } + \beta \sum_{i = 1}^N {g(x_i )w_i }
  • Linearity of Integrals
a b(αf(x)+βg(x))dx=α a bf(x)dx+β a bg(x)dx{\int_a^b {\left( {\alpha f(x)\, + \beta g(x)} \right)dx = } \alpha \int_a^b {f(x)\,dx} + \beta \int_a^b {g(x)\,dx} }
  • Can we talk about x i 2 inline? What about a bx 2dx? Inline fractions xx 2x 1x 2?

  • Big fractions

p 3(x)=(12)(x12)(x34)(x1)(1412)(1434)(141)+(12)(x12)(x34)(x1)(1412)(1434)(141)+(12)(x12)(x34)(x1)(1412)(1434)(141)+(12)(x12)(x34)(x1)(1412)(1434)(141)p_3 (x) = \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}} + \left( {\frac{1}{2}} \right)\frac{{\left( {x - \frac{1}{2}} \right)\left( {x - \frac{3}{4}} \right)\left( {x - 1} \right)}}{{\left( {\frac{1}{4} - \frac{1}{2}} \right)\left( {\frac{1}{4} - \frac{3}{4}} \right)\left( {\frac{1}{4} - 1} \right)}}
  • Diagram
P 1(Y) P 1(X) T T\begin{matrix} P_1(Y) &\to& P_1(X) \\ \downarrow &\Downarrow^\mathrlap{\sim}& \downarrow \\ T' &\to& T \end{matrix}

Will indented code work

This should be code
And this

yep

(7) A_n Quiver v1 v2 vn1 (U(k) n 1,{v i})\array{\arrayopts{\align{center}} \begin{svg} <svg xmlns="http://www.w3.org/2000/svg" width="10em" height="10em" viewBox="0 0 120 120"> <desc>A_n Quiver</desc> <g fill="none" stroke="black" stroke-width="1"> <path d="M 60 25 l 25 20 l 0 30 l -25 20 l -25 -20 l 0 -30 z" /> <g stroke-dasharray="2"> <path d="M 60 0 l 0 25" /> <path d="M 85 45 l 25 -20" /> <path d="M 85 75 l 25 20" /> <path d="M 60 95 l 0 25" /> <path d="M 35 75 l -25 20" /> <path d="M 35 45 l -25 -20" /> </g> </g> <g fill="red" stroke="none"> <circle cx="60" cy="25" r="4" /> <circle cx="85" cy="45" r="4" /> <circle cx="85" cy="75" r="4" /> <circle cx="60" cy="95" r="4" /> <circle cx="35" cy="75" r="4" /> <circle cx="35" cy="45" r="4" /> </g> <foreignObject x="45" y="0" width="16" height="20"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mn>1</mn></msub></math></foreignObject> <foreignObject x="83" y="15" width="16" height="20"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mn>2</mn></msub></math></foreignObject> <foreignObject x="5" y="25" width="20" height="25"><math xmlns="http://www.w3.org/1998/Math/MathML" display="inline"><msub><mi>v</mi><mrow><msub><mi>n</mi><mn>1</mn></msub></mrow></msub></math></foreignObject> </svg> \end{svg} } \equiv \left({U(k)}^{n_1},\{v_i\}\right)

  1. You can also refer to it as (2). Chacun à son goût!.

  2. For more information, see Wikipedia.